CF986F Oppa Funcan Style Remastered

发表于 | 阅读数

给$n$与$k$，问是否能将n分为若干个$k$的因数之和

$k$分解质因数。

  • $k.size()=0$,$NO$.
  • $k.size()=1$，$(n\mod k = 0)$
  • $k.size()=2$，$ax+by=n$
  • $k.size()>2$，$ax+by+cz…=n$ 。

复杂度$O(50(\frac{\sqrt{k}}{\log}+k^{\frac{1}{3}}\log {k^{\frac{1}{3}}}))\approx 179056941$

代码 
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#include <bits/stdc++.h>
#include <bitset>
using namespace std;
typedef long long ll;
#define pii pair<ll, ll>
#define mk make_pair
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
const int M = 4e7 + 10;

bitset<40000100> npr;
int pr[M];
int pcnt;
void Prime_init(int X)
{
    npr[1] = 1;

    for (int i = 2; i <= X; i++)
    {
        if (!npr[i])
            pr[++pcnt] = i;
        for (int j = 1; j <= pcnt && pr[j] * i <= X; j++)
        {
            npr[pr[j] * i] = 1;
            if (i % pr[j] == 0)
            {
                break;
            }
        }
    }
}
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b)
    {
        x = 1, y = 0;
        return a;
    }
    ll gd = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return gd;
}
ll cinv(ll a, ll p)
{
    ll x, y;
    exgcd(a, p, x, y);
    return (x + p) % p;
}

namespace SPFA
{
    vector<pii> g[N];
    ll dis[N];
    int vis[N];
    void spfa(int s, int c)
    {
        priority_queue<pii> q;
        for (int i = 0; i < c; i++)
            dis[i] = 2e18, vis[i] = 0;

        q.push(mk(0, s));

        vis[s] = 0;
        dis[s] = 0;
        while (!q.empty())
        {
            int x = q.top().second;
            q.pop();
            if (vis[x])
                continue;
            vis[x] = 1;
            for (pii now : g[x])
            {
                int to = now.first;
                int w = now.second;

                if (dis[to] > dis[x] + w)
                {
                    dis[to] = dis[x] + w;

                    q.push(mk(-dis[to], to));
                }
            }
        }
    }
} // namespace SPFA

using namespace SPFA;
int ans[N];
vector<int> p;

struct node
{
    ll k, n, id;
} o[N];
bool cmp(node u, node v)
{
    return u.k < v.k;
}
int main()
{

    int T;
    cin >> T;
    for (int i = 1; i <= T; i++)
    {
        ll k, n;
        cin >> n >> k;
        o[i] = (node){k, n, i};
    }
    sort(1 + o, 1 + o + T, cmp);
    vector<ll> p;
    Prime_init(M - 100);
    for (int i = 1; i <= T; i++)
    {
        if (o[i].k == o[i - 1].k)
        {
            if (p.size() == 0)
            {
                ans[o[i].id] = 0;
            }
            else if (p.size() == 1)
            {
                ans[o[i].id] = (o[i].n % p[0] == 0);
            }
            else if (p.size() == 2)
            {
                ll x = 1ll * o[i].n % p[1] * cinv(p[0], p[1]) % p[1];
                ans[o[i].id] = (x * p[0] <= o[i].n);
            }
            else
            {
                ans[o[i].id] = (dis[o[i].n % p[0]] <= o[i].n);
            }
        }
        else
        {

            ll k = o[i].k;
            p.clear();
            for (int i = 1; i <= pcnt && 1ll * pr[i] * pr[i] <= k; i++)
            {

                if (k % pr[i] == 0)
                {

                    while (k % pr[i] == 0)
                    {
                        k /= pr[i];
                    }
                    p.push_back(pr[i]);
                }
            }

            if (k > 1)
                p.push_back(k);
            if (p.size() == 0)
            {
                ans[o[i].id] = 0;
            }
            else if (p.size() == 1)
            {
                ans[o[i].id] = (o[i].n % p[0] == 0);
            }
            else if (p.size() == 2)
            {
                ll x = 1ll * o[i].n % p[1] * cinv(p[0], p[1]) % p[1];
                ans[o[i].id] = (x * p[0] <= o[i].n);
            }
            else
            {

                ll c = p[0];
                for (int j = 0; j < c; j++)
                    g[j].clear();
                for (int j = 0; j < c; j++)
                    for (int i = 1; i < p.size(); i++)
                    {
                        g[j].push_back(mk((j + p[i]) % c, p[i]));
                    }
                spfa(0, c);
                ans[o[i].id] = (dis[o[i].n % c] <= o[i].n);
            }
        }
    }
    for (int i = 1; i <= T; i++)
    {
        puts(ans[i] ? "YES" : "NO");
    }
}